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WleyPLUS WileyPLUS: MywileyPLUS I Help I Contact Us 1 Lo GENERAL ANALYTICAL PHYSICS (PHYS 211 & 2 Halliday, Fundamentals of Physics, 10e Assignment Gradebook ORION Downloadable eTextbook ent FULL SCREEN PRINTER VERSION !4BACK Chapter 22, Problem 049 A block with a mass of 13.6 g and a charge of +1.95 x 105 C is placed in an electric field with x component Ex -3.69 10 N/?, y component E,--588 N/C, and z component Ez-0. (a) what is the magnitude of the electrostatic free on the block and (b) what angle does that force make with the positive x direction? If the block is released from rest at the origin at time t-0 what are its (e) x and (d) y coordinates at t4.24 s? (a) Number (b) Number (e) Number (d) Number Units Question Attempts: 0 of 7 used SAVE FOR LATER SUBMIT ANSWER c1 202018 3he ley&.Is A Rghts Reverved. A Division of 2ahn Wiley& Sons.Inc MacBook Air esc 2 4

Explanation / Answer

Part A)

AThe magnitude of Electric field value

E^2 = (3690)^2 + (588)^2

E = 3736.55 N/C

Then apply F = qE

F = (1.95 X 10^-5)(3736.55)

F = .07286 N

Part B)

Fx = (1.95 X 10^-5)(3690) = 0.071955

Fy = (1.95 X 10^-5)(-588) = -0.011466

The angle is found using the tangent function

tan (angle) = -0.011466/0.071955

angle = -9.054 degreres (below positive x axis)

Part C)

x component of force

Fx = (1.95 X 10^-5)(3690) = 0.071955

y component of force

Fy = (1.95 X 10^-5)(-588) = -0.011466

F = ma

For x direction (.071955) = (.0136)(a)

a = 5.29 m/s^2

d = vt + .5at^2

d = (0) + (.5)(5.29)(4.24)^2

x location = 47.558 m

Part D)

For the y direction

(-.011466) = (.0136)(a)

a = -0.843 m/s^2

d = vt + .5at^2

d = (0) - (.5)(0.843)(4.24)^2

y location = -7.57835 m

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